The integral ∫01/2 ex2−x2(1−x)3/2(1+x)1/2dx is equal to

Let I=∫01/2 ex2−x2(1−x)3/2(1+x)1/2dx=∫01/2 ex1−x2+1(1−x)1-x2dx=∫01/2 ex1−x21−x+1(1−x)1−x2dxAs ddx1−x21−x=(1−x)−x1−x2+1−x2(1−x)2=−x+x2+1−x2(1−x)21−x2=1(1−x)1−x2∴I=ex1−x21−x01/2=e3/42/2−1=3e−1